Quotient Test for Convergence of Series

The quotient test can be used to determine whether a series is converging or not. The full description is in the attachment.

2. Relevant equations

3. The attempt at a solution

( i ) Why must they both follow the same behaviour? Even if p ≠ 0, it says nothing about the type of series Ʃu and Ʃv are! How can one so hastily conclude that if one converges, the other must too and if one diverges the other must too?

( ii ) If p = 0, won't it imply that the limit of un = 0 as n approaches infinity? So then series Ʃu must definitely converge. Why did they explain it the other way round (if Ʃv converges, then Ʃu must converge) Isn't Ʃu already known to be converging since limit un = 0 ?

The quotient test can be used to determine whether a series is converging or not. The full description is in the attachment.

2. Relevant equations

3. The attempt at a solution

( i ) Why must they both follow the same behaviour? Even if p ≠ 0, it says nothing about the type of series Ʃu and Ʃv are! How can one so hastily conclude that if one converges, the other must too and if one diverges the other must too?
If [itex]u_n/v_n[/itex] goes to [itex]\rho[/itex] we can say that, for n sufficiently large, [itex]u_n< (\rho+1)v_n[/itex] and, if [itex]v_n[/itex] converges, so does [itex]u_n[/itex], by the comparison test. Conversely, if [itex]u_n[/itex] diverges so does [itex]v_n[/itex]

( ii ) If p = 0, won't it imply that the limit of un = 0 as n approaches infinity? So then series Ʃu must definitely converge. Why did they explain it the other way round (if Ʃv converges, then Ʃu must converge) Isn't Ʃu already known to be converging since limit un = 0 ?

No, that does NOT imply that un goes to 0. For example, if un= n and [itex]v_n= n^2[/itex], [itex]u_n/v_n= 1/n[/itex] which goes to 0 as n goes to infinity but both un and vn go to infinity. The denominator just goes to infinity faster than the numerator.

If [itex]u_n/v_n[/itex] goes to [itex]\rho[/itex] we can say that, for n sufficiently large, [itex]u_n< (\rho+1)v_n[/itex] and, if [itex]v_n[/itex] converges, so does [itex]u_n[/itex], by the comparison test. Conversely, if [itex]u_n[/itex] diverges so does [itex]v_n[/tex]

No, that does NOT imply that un goes to 0. For example, if un= n and [itex]v_n= n^2[/itex], [itex]u_n/v_n= 1/n[/tex] which goes to 0 as n goes to infinity but both un and vn go to infinity. The denominator just goes to infinity faster than the numerator.

No, for the same reason as above. Take [itex]u_n= n^2[/itex], [itex]v_n= n[/itex].

( i ) I don't understand how you can take

[itex]u_n< (\rho+1)v_n[/itex]

and simply apply the comparison test. I tried to break it down into cases:

Case 1: 0 < p < 1

un < vn (for n tends to infinity) implying:

1a. If vn converges, so must un.1b. But if un converges, it says nothing about vn since un < vn.

2a. If un diverges, so must vn must diverge.2b. But if vn diverges, it says nothing about un since un < vn.

Case 2: p > 1
Arrives at the same paradox..

(ii)
That was a fantastic counter-example which made me realize that my thinking wasn't thorough enough. If you don't mind please tell me if my understanding is correct:

If p = 0,
It implies that vn increases at a faster rate than un such that eventually p → 0. So since vn > un for all n, then if vn converges then by comparison test un must converge!

(Would this also imply that if un diverges, then so must vn?)

(iii)
If p = ∞,

It implies that un increases at a faster rate than vn such that eventually p → ∞. So since un > vn for all n, then if vn diverges then by comparison test un must diverge!

Given [itex]\frac{u_n}{v_n}→ρ[/itex], by definition of convergence... we have for sufficiently large N, [itex]|\frac{u_n}{v_n}-ρ|<ε[/itex]
IE, [itex] (ρ-ε)v_n<u_n<(ρ+ε)v_n [/itex]
so if we pick ε<ρ for some N, we can guarantee comparison holds :-).
also, the other two problems are equivalent if you consider the reciprocal of the quotient of sequences.

Given [itex]\frac{u_n}{v_n}→ρ[/itex], by definition of convergence... we have for sufficiently large N, [itex]|\frac{u_n}{v_n}-ρ|<ε[/itex]
IE, [itex] (ρ-ε)v_n<u_n<(ρ+ε)v_n [/itex]
so if we pick ε<ρ for some N, we can guarantee comparison holds :-).
also, the other two problems are equivalent if you consider the reciprocal of the quotient of sequences.

I'm sorry but i dont really understand what you just wrote.. first of all what is ε?

Here is the definition of convergence for a sequence.
in order for [itex]a_n→a[/itex] it follows that for all ε>0 there exists some N so that for natural numbers n>N, [itex]|a_n-a|<ε[/itex]
that is... as we get further out into our sequence, the behavior we'd see is that the sequence will get closer to it's limit, or the distance between the limit and the sequence will get smaller and smaller.
take [itex]|\frac{1}{n}-0|<\epsilon[/itex]... since 1/n→0,
that means we can pick any interval and find an N so that it works for all n>N
for instance... if we wanted to consider the case for ε=1/100...
[itex]\frac{1}{n}<\frac{1}{100}⇔n>100=N[/itex]
the same way we can do it with the quotient converging to ρ. Halls of Ivy specifically picked for ε=1 in this case.
I suggest though keeping ε<ρ, for positive sequences (makes it a lil prettier)

Here is the definition of convergence for a sequence.
in order for [itex]a_n→a[/itex] it follows that for all ε>0 there exists some N so that for natural numbers n>N, [itex]|a_n-a|<ε[/itex]
that is... as we get further out into our sequence, the behavior we'd see is that the sequence will get closer to it's limit, or the distance between the limit and the sequence will get smaller and smaller.
take [itex]|\frac{1}{n}-0|<\epsilon[/itex]... since 1/n→0,
that means we can pick any interval and find an N so that it works for all n>N
for instance... if we wanted to consider the case for ε=1/100...
[itex]\frac{1}{n}<\frac{1}{100}⇔n>100=N[/itex]
the same way we can do it with the quotient converging to ρ. Halls of Ivy specifically picked for ε=1 in this case.
I suggest though keeping ε<ρ, for positive sequences (makes it a lil prettier)

I get it now, thank you! But I don't see any relation in explaining how the inequality post explains why as long as p≠0 both series will be share the same behaviour..